Theorem iunssf

Description: Subset theorem for an indexed union. (Contributed by Glauco Siliprandi, 3-Mar-2021)

	Ref	Expression
Hypothesis	iunssf.1	⊢ Ⅎ 𝑥 𝐶
Assertion	iunssf	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )

Proof

Step	Hyp	Ref	Expression
1		iunssf.1	⊢ Ⅎ 𝑥 𝐶
2		df-iun	⊢ ∪ 𝑥 ∈ 𝐴 𝐵 = { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 }
3	2	sseq1i	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 } ⊆ 𝐶 )
4		abss	⊢ ( { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 } ⊆ 𝐶 ↔ ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
5		df-ss	⊢ ( 𝐵 ⊆ 𝐶 ↔ ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
6	5	ralbii	⊢ ( ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
7		ralcom4	⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑦 ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
8	1	nfcri	⊢ Ⅎ 𝑥 𝑦 ∈ 𝐶
9	8	r19.23	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
10	9	albii	⊢ ( ∀ 𝑦 ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
11	6 7 10	3bitrri	⊢ ( ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )
12	3 4 11	3bitri	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )